-2
-1
\(-\frac{1}{2}\)
1
Correct answer is D
\(\frac{x^2 + x - 2}{2x^2 + x - 3}\)
= \(\frac{x^2 + 2x - x - 2}{2x^2 + 3x - 2x - 3}\)
= \(\frac{x(x + 2) - 1(x + 2)}{x(2x + 3) - 1(2x + 3)}\)
= \(\frac{(x - 1)(x + 2)}{(x - 1)(2x + 3)}\)
= \(\frac{x + 2}{2x + 3}\)
At x = -1,
= \(\frac{-1 + 2}{2(-1) + 3}\)
= \(\frac{1}{1}\)
= 1